Is it possible to inscribe a square in an arbitrary convex quadrilateral $ABCD$ with only compass and straight edge?
I know how to construct a square inscribed in a triangle, but I don't know how to do this.
Inscribe a square in an arbitrary quadrilateral using compass and straight edge
euclidean-geometrygeometryplane-geometry
Related Solutions
I can describe how I will do it with Dr. Geo, a free interactive geometry software I am developing.
- From the menu Edit>Customise interface, you select the tools you want: select and move, point, circle, segment and macro tools,
Select only the tools you want
- Then you save your empty sketch under the name Minimimal for example,
- Next time you load the Minimal sketch its comes with your customized UI, including the tools to register and play macro-construction.
Geometric canvas with minimal tools
I use this feature to limit the tools used for primary school activities.
Well, the side length of a regular hexagon equals the circumradius, so you may simply draw an inscribed regular hexagon and rotate it until one of its sides goes through the given point. Of course this can be done only if the distance of the given point from the center of the circle is $\geq \frac{\sqrt{3}}{2}R$.
Let $P$ be the given point and $O$ the center$^{(*)}$ of the given circle $\Gamma$. You may
- Take any $A\in\Gamma$ and construct $B$ as one of the intersections between $\Gamma$ and a circle centered at $A$ with radius $AO$ (now $AB$ equals the radius)
- Construct $C$ as the intersection between $AB$ and a circle centered at $O$ through $P$
- Construct $D$ as one of the intersections between $\Gamma$ and a circle centered at $P$ with radius $AC$.
$DP$ meets $\Gamma$ again at $E$ and $DE$ is a solution.
$(*)$ If $O$ is not previously drawn, you may simply construct it as the intersection of the axis of two different chords. Here an alternative construction which exploits the intersecting chords theorem:
- Let $QR$ the diameter of $\Gamma$ through $P$, with $Q$ being closer to $P$ than $R$;
- Let $S$ be one of the intersections between $\Gamma$ and the perpendicular to $OP$ through $P$ ($PS$ is the geometric mean of $PQ$ and $PR$)
- Let $M$ be the midpoint of $OQ$ and $T$ be the intersection, closer to $S$, between the circle centered at $M$ through $O$ and the parallel to $OP$ through $S$
- Let $U$ be the projection of $T$ on $OP$: now the geometric mean of $UQ,UO$ is also the geometric mean of $PQ,PR$ and the arithmetic mean of $UQ,UO$ is half the arithmetic mean of $PQ,PR$, so...
- Draw a circle centered at $P$ with radius $UQ$ and let $V$ be one of the intersections with $\Gamma$.
If $VP$ meets $\Gamma$ again at $W$, $VW$ is a chord whose length equals $OQ$.
This construction is a bit more involved than the previous one, but both approaches can be easily modified in order to solve the problem of drawing a chord through $P$ of any allowed length.
Best Answer
Here’s a simple construction for the interesting case where the square must touch all four sides. (The other cases are easily constructed by starting with a square with three vertices on two sides of $ABCD$, and scaling it so its fourth vertex touches the third side of $ABCD$ using a homothety about the intersection of the initial two sides.)
Construct $E$ on $\overrightarrow{DA}$ with $\angle DCE = 45^\circ$, and $F$ on $\overrightarrow{CB}$ with $\angle CDF = 45^\circ$. Draw $CG$ and $DH$ perpendicular to $CD$ with $CD \parallel EG \parallel FH$. Let $GH$ intersect $AB$ at $I$. Then $I$ is a vertex of the square.
Drop perpendicular $IJ$ to $CD$. Construct $K$ on $\overrightarrow{DA}$ with $\angle DJK = 45^\circ$, and $L$ on $\overrightarrow{CB}$ with $\angle CJL = 45^\circ$. Then $K, L$ are two more vertices of the square and the last one can be constructed.